Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical and the nodes have the same value. 给定两个二叉树,判断是否完全相同. 若两个二叉树完全相同,则每个节点的左子树和右子树必然都相同. 可以采用递归方式,从根节点开始:…

Input: 1 1 / \ / \ 2 3 2 3 [1,2,3], [1,2,3] Output: true Example 2: Input: 1 1 / \ 2 2 [1,2], [1,null,2] Output: false Example 3: Input: 1 1 / \ / \ 2 1 1 2 [1,2,1], [1,1,2] Output: false 判断两个二叉树是否相等. 一:使用递归. public boolean isSameTree(TreeNode p, Tre…

Given two binary trees, write a function to check if they are the same or not. Two binary trees are considered the same if they are structurally identical and the nodes have the same value. Example 1: Input: / \ / \ [,,], [,,] Output: true Example 2:…

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then…

迭代版本用的是二叉树的DFS,中的root->right->left ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 给定两个二叉树,写一个函数判断他们是否是相同的. 如果两个二叉树的结构相同而且每个节点里面的值也相同,那么认为他们是相同的二叉树. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++…

Invert a binary tree 翻转一棵二叉树 假设有如下一棵二叉树: 4  / \   2    7  / \   / \ 1  3 6  9翻转后: 4     /    \    7     2   / \    / \  9  6  3  1 这里采用递归的方法来处理.遍历结点,将每个结点的两个子结点交换位置即可. 从左子树开始,层层深入,由底向上处理结点的左右子结点:然后再处理右子树 全部代码如下: public class InvertBinaryTree { public…

/* Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, t…

翻译 给定一个二叉树.返回其兴许遍历的节点的值. 比如: 给定二叉树为 {1. #, 2, 3} 1 \ 2 / 3 返回 [3, 2, 1] 备注:用递归是微不足道的,你能够用迭代来完毕它吗? 原文 Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: R…

Level:   Easy 题目描述: Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge them into a new binary tree. The merge rule is that if two…

[145-Binary Tree Postorder Traversal(二叉树非递归后序遍历)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive sol…